Question

A baton twirler has a baton of length 0.36 m with masses of 0.48 kg at each end. Assume the rod itself is massless. The rod is first rotated about its midpoint. It is then rotated about one of its ends, and in both cases uniformly accelerates from 0 rad/s to 2.4 rad/s in 3.6 s. Find the torque exerted by the twirler on the baton when it is rotated about its middle. Find the torque exerted by the twirler on the baton when it is rotated about its end.

Answer 1

Answer:

Part a)

When rotated about the mid point

$\tau = 0.021Nm$

Part b)

When rotated about its one end

$\tau = 0.042 Nm$

Explanation:

As we know that the angular acceleration of the rod is rate of change in angular speed

so we will have

$\alpha = \frac{\Delta \omega}{\Delta t}$

$\alpha = \frac{2.4 - 0}{3.6}$

$\alpha = 0.67 rad/s^2$

Part a)

When rotated about the mid point

$I = 2mr^2$

$I = 2(0.48)(0.18)^2$

$I = 0.0311 kg m^2$

now torque is given as

$\tau = 0.0311 (0.67)$

$\tau = 0.021Nm$

Part b)

When rotated about its one end

$I = m(2r)^2$

$I = (0.48)(0.36)^2$

$I = 0.0622 kg m^2$

now torque is given as

$\tau = (0.0622)(0.67)$

$\tau = 0.042 Nm$

Answer 2

When the rod is rotated about the mid point, the torque is 0.021 Nm.

When the rod is rotated about its one end, the torque is 0.041 Nm.

The angular acceleration of the rod is calculated as follows;

$\alpha = \frac{\Delta \omega }{\Delta t} = \frac{2.4}{3.6} = 0.67 \ rad/s^2$

When the rod is rotated about the mid point, the moment of inertia is calculated as;

$I = 2mr^2\\\\I = 2(0.48)(\frac{0.36}{2} )^2\\\\I = 0.031 \ kgm^2$

The torque is calculated as follows;

$\tau = I \alpha \\\\\tau = (0.031) (0.67)\\\\\tau = 0.021 \ Nm$

When the rod is rotated about its one end, the moment of inertia is calculated as follows;

$I = ML^2\\\\I = 0.48 \times 0.36^2\\\\I = 0.062 \ kgm^2$

The new torque is calculated as follows;

$\tau = (0.062) (0.67)\\\\\tau = 0.041 \ Nm$

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